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EPFL

IMAC-IS-ENAC

Cours de Dynamique des structures

Prof. I. Smith

EXERCICE 5

Corrig

CORRIGE EXERCICE No 5

M

h

P (t)

t

P (t)

I I

t 1

P0

rigidit :m

N

h

EIK

6

310016.2

122 =

=

pulsation propre: n = KM

20.08 rad/s

Lintgrale de Duhamel scrit (pas damortissement) :

=t

n

n

dtPM

tX

0

))(sin()(1

)(

avec: P( ) = P0 1 - t 1

0 t 1

P( ) = 0 t1

0 t t 1 :

X(t) =P 0

M n1-

t 1sin n (t- ) d

= 0

= t

=P 0

M nsin n (t- ) d

= 0

= t

-P0

M n t 1 sin n (t- ) d

= 0

= t

Rappel : Intgration par partie

= dvuvudvu )()(')()()(')(

avec1)('

)(

=

=

u

u

n

n

n

tv

tv

))(cos()(

))(sin()('

=

=

X(t) =P0

M n2cos n (t- ) = 0

= t-

P 0

M n2 t 1 cos n (t- ) = 0

= t- cos n (t- ) d

=0

= t

X(t) = P0

M n21-cos n t - P0

M n2 t 1t + 1

nsin n (t- )

= 0 = t

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EPFL

IMAC-IS-ENAC

Cours de Dynamique des structures

Prof. I. Smith

EXERCICE 5

Corrig

X(t) =P0

M n21- cos( n t) - t

t 1+ 1

n t 1sin( n t)

+= )cos(11

)sin()(11

0 ttt

tK

P

tX nnn &

t t 1 :

Le systme est un oscillateur harmonique dont les conditions initiales sont:

X0 = X( t 1 ) =P0

K-cos( n t 1 ) + 1

n t 1sin( n t 1 )

+== )cos(

11

)sin()(11

0

10t

ttt

K

PtXX

nnn

&&

Rappel oscillateur harmonique : )cos()sin()( tBtAtxnn

+=

On trouve donc : )cos()sin()( 00

tXtX

tXnn

n

+=&

avec 1ttt =

dplacement statique : stat =P0

K 4,96 cm

en t = t1:X(t 1 ) -0,287 stat

s

mtX 906.0)( 1 &

dplacement maximum : sttX 15.00)( =&

cmXstat

176.985.1max =

Rponse

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1

0.00 0.50 1.00 1.50 2.00 2.50

temps [s]

dpl.[m]